A star producing light of wavelength 6000 ma | vedclass

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Question

(A) The Doppler shift in wavelength $(\Delta \lambda)$ for a source moving away from the observer is given by the formula: $\Delta \lambda = \frac{v \

Vedclass · Accepted Answer

$0.1$

Vedclass · Answer

(A) The Doppler shift in wavelength $(\Delta \lambda)$ for a source moving away from the observer is given by the formula: $\Delta \lambda = \frac{v \lambda}{c}$.Given:Speed of the star $(v)$ = $5 \ km/s = 5000 \ m/s$.Wavelength $(\lambda)$ = $6000 \ \mathring{A}$.Speed of light $(c)$ = $3 	imes 10^8 \ m/s$.Substituting these values into the formula:$\Delta \lambda = \frac{5000 	imes 6000}{3 	imes 10^8} \ \mathring{A}$.$\Delta \lambda = \frac{3 	imes 10^7}{3 	imes 10^8} \ \mathring{A} = 0.1 \ \mathring{A}$.Therefore, the shift in wavelength is $0.1 \ \mathring{A}$.

$A$ star producing light of wavelength $6000 \ \mathring{A}$ moves away from the Earth with a speed of $5 \ km/s$. Due to the Doppler effect, the shift in wavelength will be $...... \ \mathring{A}$ $(c = 3 \times 10^8 \ m/s)$.

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